2 * Copyright (c) 1992, 1993
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5 * This software was developed by the Computer Systems Engineering group
6 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
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33 * $FreeBSD: src/sys/libkern/qdivrem.c,v 1.8 1999/08/28 00:46:35 peter Exp $
34 * $DragonFly: src/sys/libkern/qdivrem.c,v 1.4 2004/01/26 11:09:44 joerg Exp $
38 * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
39 * section 4.3.1, pp. 257--259.
42 #include <libkern/quad.h>
44 #define B (1 << HALF_BITS) /* digit base */
46 /* Combine two `digits' to make a single two-digit number. */
47 #define COMBINE(a, b) (((u_long)(a) << HALF_BITS) | (b))
49 /* select a type for digits in base B: use unsigned short if they fit */
50 #if ULONG_MAX == 0xffffffff && USHRT_MAX >= 0xffff
51 typedef unsigned short digit;
57 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
58 * `fall out' the left (there never will be any such anyway).
59 * We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS.
62 shl(digit *p, int len, int sh)
66 for (i = 0; i < len; i++)
67 p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
68 p[i] = LHALF(p[i] << sh);
72 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
74 * We do this in base 2-sup-HALF_BITS, so that all intermediate products
75 * fit within u_long. As a consequence, the maximum length dividend and
76 * divisor are 4 `digits' in this base (they are shorter if they have
80 __qdivrem(u_quad_t uq, u_quad_t vq, u_quad_t *arq)
87 digit uspace[5], vspace[5], qspace[5];
90 * Take care of special cases: divide by zero, and u < v.
94 static volatile const unsigned int zero = 0;
96 tmp.ul[H] = tmp.ul[L] = 1 / zero;
111 * Break dividend and divisor into digits in base B, then
112 * count leading zeros to determine m and n. When done, we
114 * u = (u[1]u[2]...u[m+n]) sub B
115 * v = (v[1]v[2]...v[n]) sub B
117 * 1 < n <= 4 (if n = 1, we use a different division algorithm)
118 * m >= 0 (otherwise u < v, which we already checked)
125 u[1] = HHALF(tmp.ul[H]);
126 u[2] = LHALF(tmp.ul[H]);
127 u[3] = HHALF(tmp.ul[L]);
128 u[4] = LHALF(tmp.ul[L]);
130 v[1] = HHALF(tmp.ul[H]);
131 v[2] = LHALF(tmp.ul[H]);
132 v[3] = HHALF(tmp.ul[L]);
133 v[4] = LHALF(tmp.ul[L]);
134 for (n = 4; v[1] == 0; v++) {
136 u_long rbj; /* r*B+u[j] (not root boy jim) */
137 digit q1, q2, q3, q4;
140 * Change of plan, per exercise 16.
143 * q[j] = floor((r*B + u[j]) / v),
144 * r = (r*B + u[j]) % v;
145 * We unroll this completely here.
147 t = v[2]; /* nonzero, by definition */
149 rbj = COMBINE(u[1] % t, u[2]);
151 rbj = COMBINE(rbj % t, u[3]);
153 rbj = COMBINE(rbj % t, u[4]);
157 tmp.ul[H] = COMBINE(q1, q2);
158 tmp.ul[L] = COMBINE(q3, q4);
164 * By adjusting q once we determine m, we can guarantee that
165 * there is a complete four-digit quotient at &qspace[1] when
168 for (m = 4 - n; u[1] == 0; u++)
170 for (i = 4 - m; --i >= 0;)
175 * Here we run Program D, translated from MIX to C and acquiring
176 * a few minor changes.
178 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
181 for (t = v[1]; t < B / 2; t <<= 1)
184 shl(&u[0], m + n, d); /* u <<= d */
185 shl(&v[1], n - 1, d); /* v <<= d */
191 v1 = v[1]; /* for D3 -- note that v[1..n] are constant */
192 v2 = v[2]; /* for D3 */
197 * D3: Calculate qhat (\^q, in TeX notation).
198 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
199 * let rhat = (u[j]*B + u[j+1]) mod v[1].
200 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
201 * decrement qhat and increase rhat correspondingly.
202 * Note that if rhat >= B, v[2]*qhat < rhat*B.
204 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
205 uj1 = u[j + 1]; /* for D3 only */
206 uj2 = u[j + 2]; /* for D3 only */
212 u_long nn = COMBINE(uj0, uj1);
216 while (v2 * qhat > COMBINE(rhat, uj2)) {
219 if ((rhat += v1) >= B)
223 * D4: Multiply and subtract.
224 * The variable `t' holds any borrows across the loop.
225 * We split this up so that we do not require v[0] = 0,
226 * and to eliminate a final special case.
228 for (t = 0, i = n; i > 0; i--) {
229 t = u[i + j] - v[i] * qhat - t;
231 t = (B - HHALF(t)) & (B - 1);
236 * D5: test remainder.
237 * There is a borrow if and only if HHALF(t) is nonzero;
238 * in that (rare) case, qhat was too large (by exactly 1).
239 * Fix it by adding v[1..n] to u[j..j+n].
243 for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
244 t += u[i + j] + v[i];
248 u[j] = LHALF(u[j] + t);
251 } while (++j <= m); /* D7: loop on j. */
254 * If caller wants the remainder, we have to calculate it as
255 * u[m..m+n] >> d (this is at most n digits and thus fits in
256 * u[m+1..m+n], but we may need more source digits).
260 for (i = m + n; i > m; --i)
262 LHALF(u[i - 1] << (HALF_BITS - d));
265 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
266 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
270 tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
271 tmp.ul[L] = COMBINE(qspace[3], qspace[4]);